SOLUTION: 8sinx-cosx =4

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Question 1074030: 8sinx-cosx =4
Answer by ikleyn(52905) About Me  (Show Source):
You can put this solution on YOUR website!
.
8sinx - cosx = 4.
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Introduce new variable y = cos(x).

Then your equation becomes

8%2Asqrt%281-y%5E2%29 = 4 + y.

Square both sides.

64(1-y^2) = 16 + 8y + y^2,

65y^2 + 8y - 48 = 0.

y%5B1%2C2%5D = %28-8+%2B-+sqrt%28%28-8%29%5E2+%2B+4%2A64%2A48%29%29%2F%282%2A65%29 = %28-8+%2B-+112%29%2F130.


1.  y%5B1%5D = -120%2F130 = -12%2F13  --->  cos(x) = -12%2F13  --->  x = +/- arccos%28-12%2F13%29 + 2k%2Api, k = 0, +/-1, +/-2, . . . 


    Check by substituting x into the original equation. 

             Only the solution in the second quadrant works: x = + arccos%28-12%2F13%29 + 2k%2Api, k = 0, +/-1, +/-2, . . . 

             The solution in the third quadrant, x = - arccos%28-12%2F13%29 + 2k%2Api, k = 0, +/-1, +/-2, . . . doesn't work (is extraneous).



2.  y%5B2%5D = 104%2F130 = 4%2F5  --->  cos(x) = {{4/5}}}  --->  x = +/- arccos%284%2F5%29 + 2k%2Api, k = 0, +/-1, +/-2, . . . 


    Check by substituting x into the original equation. 

             Only the solution in the first quadrant works: x = + arccos%284%2F5%29 + 2k%2Api, k = 0, +/-1, +/-2, . . . 

             The solution in the fourth quadrant, x = - arccos%284%2F15%29 + 2k%2Api, k = 0, +/-1, +/-2, . . . doesn't work (is extraneous).


Answer.  The solutions are these angles 

         arccos%28-12%2F13%29 + 2k%2Api, k = 0, +/-1, +/-2, . . .   and   arccos%284%2F5%29 + 2k%2Api, k = 0, +/-1, +/-2, . . .




Plots y = 8sin(x) - cos(x) (red) and y = 4 (green)